research38 min studyLesson 827 in Physics

Unruh detectors and acceleration temperature

Derive the thermal detailed-balance ratio for a uniformly accelerated detector while keeping its state, trajectory, and switching assumptions explicit.

Before you begin

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The question for today

What you’ll understand

Define detector response through a field correlation function.
Compute a stationary excitation-to-deexcitation ratio.
Separate detector thermality from an observer-independent thermal stress tensor.

A particle definition becomes operational when an apparatus is specified. An Unruh–DeWitt detector is an idealized localized quantum system whose internal transitions couple to a field along a prescribed worldline. Its response depends jointly on the quantum state, the trajectory, and the switching of the interaction. Uniform acceleration through the Minkowski vacuum gives a particularly sharp result: in the stationary long-interaction limit, transition rates obey thermal detailed balance. That statement is more precise than saying that acceleration fills otherwise empty space with ordinary hot matter.

Correlations along a worldline determine the response

Take a two-level detector with positive angular-frequency gap Ω and energy gap ħΩ. At leading nontrivial order in a weak linear coupling, its excitation probability is proportional to the Fourier transform of the field's Wightman two-point function pulled back to the worldline. A smooth switching function controls when the coupling acts. The omitted proportionality factor includes the squared coupling and the detector's internal matrix element; neither changes the detailed-balance ratio when the same transition is compared in both directions.

Use signature (−,+,+,+). In the response equations set ħ=c=1 and use proper time τ, so Ω and proper acceleration a have inverse-length units. For uniform acceleration in Minkowski spacetime, the worldline has t=a⁻¹sinh(aτ) and x=a⁻¹cosh(aτ). The massless scalar vacuum correlation on this worldline depends only on the proper-time difference. Its imaginary-time structure supplies the thermal frequency relation. A different field, boundary, state, trajectory, or coupling must be analyzed on its own terms.

\mathcal F(\Omega)=\int d\tau\,d\tau'\,\chi(\tau)\chi(\tau')e^{-i\Omega(\tau-\tau')}W(x(\tau),x(\tau')),\qquad W(s)=-\frac{a^2}{16\pi^2\sinh^2[\tfrac a2(s-i\epsilon)]}

χ is a smooth real switching function and W is the Minkowski-vacuum Wightman function for a free massless scalar in four dimensions, with its distributional iε prescription. The displayed stationary W uses the uniform-acceleration trajectory and s=τ−τ′; a detector is treated perturbatively.↔ Scroll wide equations horizontally.

Detailed balance defines the acceleration temperature

In the stationary long-duration limit, the excitation response per unit proper time has a Planck denominator. Deexcitation also contains spontaneous emission. Their ratio, rather than either rate alone, fixes the temperature through detailed balance. Restoring constants gives kB T=ħ a/(2π) when a is geometric inverse length and c=1; in SI units T=ħ aSI/(2π c kB). The acceleration is the detector's proper acceleration, not a coordinate second derivative.

A weakly coupled two-level detector that equilibrates has excited-to-ground population ratio exp(−ħΩ/kB T). Its excited probability is therefore a normalized two-state Boltzmann probability. It is not the Bose mean occupation appearing in the field-response denominator. Confusing these quantities can yield an impossible excited population greater than one. The rate ratio is independent of an overall coupling strength, whereas the time required to approach equilibrium is not.

\dot{\mathcal F}(\Omega)=\frac{\Omega}{2\pi(e^{2\pi\Omega/a}-1)},\qquad \frac{R_\uparrow}{R_\downarrow}=e^{-2\pi\Omega/a},\qquad P_e=\frac{1}{1+e^{2\pi\Omega/a}}

The formula uses Ω>0, a>0, a free massless scalar, the Minkowski vacuum, stationary uniform acceleration, and the long-interaction limit. R↑ and R↓ include the same detector coupling factor; Pe additionally assumes weak-coupling equilibration of the two-level system.↔ Scroll wide equations horizontally.

Thermal response is a conditional operational statement

The Minkowski vacuum has vanishing renormalized stress tensor in its standard normalization, despite the accelerated detector's thermal response. Maintaining the acceleration requires an external agent. Energy accounting includes that agent and the detector-field interaction, so the response does not imply a pre-existing material bath with observer-independent local energy density. An inertial detector with eternal stationary coupling in the same vacuum has no positive-gap excitation rate.

Finite switching produces transients, including possible excitations of an inertial detector. Abrupt idealizations require regularization and can create artifacts. A slowly changing acceleration may admit useful approximations, but arbitrary nonuniform motion does not obey the exact uniform-acceleration formula with an instantaneous substitution. The distinction matters near black holes: a hovering apparatus and a freely falling apparatus follow different worldlines, while the field state must still be specified. The Unruh calculation explains a local operational mechanism without solving an evaporating black-hole spacetime.

Work it through

Separate a rate ratio from a two-state population

A weakly coupled detector has stationary uniform acceleration chosen so that 2πΩ/a=ln 3. Assume the stated long-interaction scalar-vacuum response and eventual two-level equilibration. Find the upward-to-downward rate ratio, both response rates with their common detector factor removed, and the excited population.

The Boltzmann rate ratio is $R_\uparrow/R_\downarrow=e^{-\ln3}=1/3$ . This number compares transition directions and is not itself the excited population.
The excitation response is $\dot{\mathcal F}(\Omega)=\Omega/[2\pi(3-1)]=\Omega/(4\pi)$ . Detailed balance gives the deexcitation response $3\Omega/(4\pi)$ .
Equilibrium requires $P_e/P_g=1/3$ and $P_e+P_g=1$ . Therefore Pe=1/4 and Pg=3/4, independently of the common weak coupling factor.

The result

The rate ratio is 1/3, the normalized responses are Ω/(4π) and 3Ω/(4π), and the equilibrium excited population is 1/4.

The living notebook

Explore the model

Interactive lab

Interactive lab · markov

A rate ratio and a relaxation time carry different information

For each fresh comparison, Reset and set BOTH Transition p and Transition q. Compare (p,q)=(0.10,0.30) and (0.20,0.60). Both have stationary P1=0.25. Starting from state 0, take exactly two Steps after each Reset: predict P1=0.16 and 0.24. Each Step is one transition iteration. This discrete two-state probability model illustrates equal equilibrium ratios with different relaxation speeds; it does not calculate detector response, acceleration, quantum-field correlations, or an Unruh temperature.

P(state 0)P(state 1)Stationary distribution

P(state 1) at step 0: 0%
Stationary P(state 1): 60%
Chain behavior: Monotone convergence

Transition p · 0 → 10.3Transition q · 1 → 00.2

t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn equilibrium probability can remain fixed while the approach to equilibrium changes.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Stationary population balance gives Pe R↓=Pg R↑ with Pe+Pg=1, so Pe=R↑/(R↑+R↓). This differs from the field Bose factor and from the unnormalized rate ratio.

Your turn

Try it yourself

Keep the detector gap fixed but double the uniform proper acceleration from the worked example. Assuming the same stationary weak-coupling equilibration conditions, compute the new detailed-balance ratio and excited population. Use √3≈1.732. Explain why the population remains below one half.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Doubling a halves the exponent 2πΩ/a. Normalize the new excited-to-ground ratio rather than identifying it with Pe.

Compare with the solution

The new exponent is (ln 3)/2, giving $R_\uparrow/R_\downarrow=1/\sqrt3\simeq0.57735$ . The population is $P_e=1/(1+\sqrt3)\simeq0.36603$ , with Pg≈0.63397. Both are normalized and the excitation has increased from 0.25. For a positive gap at any finite positive temperature, the Boltzmann factor remains below one, so Pe remains below one half; it tends toward one half only as the acceleration temperature grows without bound within this idealized detector model.

Keep the idea

Take it with you

Detector response is a correlation-function calculation with specified switching and trajectory.
Uniform acceleration gives thermal detailed balance under stationary long-interaction assumptions.
A two-level population, a field occupation, and a renormalized stress tensor are different observables.

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